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Chapter 2: Problem 191

A particle moves in a straight line with an acceleration \(\left(12 s^{2}\right) \mathrm{ms}^{-2}\) where \(s\) is the displacement of the particle in metre from \(O\), a fixed point on the line, at time \(t\) seconds. The particle has zero velocity when its displacement from \(O\) is \(-2 \mathrm{~m}\). Find the velocity (in \(\mathrm{m} / \mathrm{s}\) ) of the particle as it passes through \(O .\)

Short Answer

Expert verified
The velocity of the particle as it passes through point O is \(0\: m/s\).

Step by step solution

01

Find the initial velocity (u) when displacement is -2 meters

Given acceleration, \(a = 12s^2 m/s^2\) and the fact that the particle has zero velocity when its displacement is -2 meters, we can find its initial velocity. From the first formula, the velocity equation is: \(v = u + at\) Since v is 0 when the displacement (s) is -2 meters, we plug it in the equation: \(0 = u + a(-2)\) Now substitute a with the given acceleration (12s^2): \(0 = u-24s^2\) We now have the equation of initial velocity (u) in terms of displacement (s).
02

Find the time (t) it takes for the particle to pass through point O

The particle passes through point O when the displacement s is 0 meters. Using our derived equation for initial velocity: \(0 = u - 24(0)^2\) We now have: \(0 = u\) The initial velocity (u) is equal to 0.
03

Find the velocity (v) of the particle as it passes through point O

Now, using the given displacement equation: \(s = ut + \frac{1}{2}at^2\) Since our displacement (s) is 0 and initial velocity (u) is 0, \(0 = 0*t + \frac{1}{2}(12s^2)t^2\) Now, we can find the time (t) it takes for the particle to reach point O: \(0 = (6s^2)t^2\) Since the particle does pass through point O, t is not equal to 0. Thus, for the equation to hold true, the displacement (s) must be \(0\; when\; t e 0\). Now, with the derived equation: velocity (v) as a function of acceleration (a) and time (t) is: \(v(u=0) = at\) By substitution, we get: \(v = (12s^2)t\) For the particle to cross point O, the displacement (s) must be 0: \(v(0) = (12(0)^2)t\) Therefore, the velocity (v) of the particle as it passes through point O is \(0\: m/ss\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acceleration
Acceleration is a fundamental concept in motion in one dimension. It describes how the velocity of an object changes over time. In this particular problem, the acceleration is given as a function of displacement, specifically as \(a = 12s^2\) meters per second squared. This means that the acceleration depends on how far the particle is from a fixed point \(O\). Understanding that acceleration can vary with position helps in solving problems where constant acceleration formulas do not apply. Here, acceleration isn't constant; it's a function of displacement \(s\). When dealing with such problems:
  • Identify how acceleration changes with displacement or time.
  • Understand how a change in displacement (\(s\)) impacts acceleration (\(a\)).
This knowledge helps you set up equations to find other unknown variables like velocity or time.
Displacement
Displacement measures how far the particle is from a specific point, \(O\), in a specified direction. It is a vector quantity, meaning it has both magnitude and direction. In this exercise, you are given that when the displacement from \(O\) is -2 meters, the velocity of the particle is zero. Displacement plays a critical role in determining motion metrics such as velocity and acceleration.When analyzing the motion of particles:
  • Use given displacement values to determine changes in motion, like initial conditions or boundary conditions.
  • Realize how displacement links directly with the velocity function as it initiates from a point with a particular velocity condition.
Knowing the role of displacement in the calculated expressions will help solve for velocity or time.
Velocity
Velocity is the rate at which the displacement changes with time. It's another vector quantity, meaning it includes direction, making it distinct from speed, which is scalar. Initially, the problem states that the particle has zero velocity when its displacement from \(O\) is -2 meters. Velocity is integral in finding how fast the particle is moving at any given point in its path.Key ideas when working with velocity:
  • Recognize initial conditions where velocity can be zero, helping determine its state at different positions.
  • Understand how velocity and displacement are interconnected through functions derived from acceleration.
  • Use given equations to assess velocity at specific points (e.g., as it passes through point \(O\)).
In this exercise, the problem derives velocity by identifying the changes in acceleration and corresponding displacement, helping find the velocity through point \(O\).

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Most popular questions from this chapter

Rain is falling vertically with a speed of \(30 \mathrm{~ms}^{-1}\). A woman rides a bicycle at a speed of \(10 \mathrm{~ms}^{-1}\) in the north to south direction. What is the direction in which she should hold her umbrella? \(\begin{array}{ll}\text { (A) } \theta=\tan ^{-1} 3 & \text { (B) } \theta=\tan ^{-11}\end{array}\) (C) \(\theta=\tan ^{-1} \frac{2}{3}\) (D) \(\theta=\tan ^{-1} \frac{3}{2}\)

A glass wind screen whose inclination with the vertical can be changed is mounted on a car. The car moves horizontally with a speed of \(2 \mathrm{~m} / \mathrm{s}\). At what angle \(\alpha\) with the vertical should the wind screen be placed so that rain drops falling vertically downwards with velocity \(6 \mathrm{~m} / \mathrm{s}\) strike the wind screen perpendicularly? (A) \(\tan ^{-1}\left(\frac{1}{3}\right)\) (B) \(\tan ^{-1}(3)\) (C) \(\cos ^{-1}(3)\) (D) \(\sin ^{-1}\left(\frac{1}{3}\right)\)

A \(2 \mathrm{~m}\) wide truck is moving with a uniform speed \(v_{0}=\) \(8 \mathrm{~m} / \mathrm{s}\) along a straight horizontal road. A pedestrian starts to cross the road with a uniform speed \(v\) when the truck is \(4 \mathrm{~m}\) away from him. The minimum value of \(v\) so that he can cross the road safely is (A) \(2.62 \mathrm{~m} / \mathrm{s}\) (B) \(4.6 \mathrm{~m} / \mathrm{s}\) (C) \(3.57 \mathrm{~m} / \mathrm{s}\) (D) \(1.414 \mathrm{~m} / \mathrm{s}\)

In a harbour, wind is blowing at the speed of \(72 \mathrm{~km} / \mathrm{h}\) and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of \(51 \mathrm{~km} / \mathrm{h}\) to the north, what is the direction of the flag on the mast of the boat? (A) \(\tan ^{-1} \frac{51}{72 \sqrt{2}-51}\) (B) \(\tan ^{-1} \frac{72 \sqrt{2}-51}{51}\) (C) \(\tan ^{-1} 1\) (D) None

A bullet is fired with a gun from a tower horizontally with a velocity \(400 \mathrm{~m} / \mathrm{s}\). At the same time, a stone is dropped from the same tower. (A) The stone will reach the ground first (B) The bullet will reach the ground first (C) Both will reach the ground at the same time (D) (A) and (B) according to the height of tower
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